The equation of a circle $C$ is $x^2+y^2+4x-4y-41 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+4x) + (y^2-4y) = 41$ $(x^2+4x+4) + (y^2-4y+4) = 41 + 4 + 4$ $(x+2)^{2} + (y-2)^{2} = 49 = 7^2$ Thus, $(h, k) = (-2, 2)$ and $r = 7$.